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RSB [31]
3 years ago
11

A family wants to build a rectangular garden on one side of a barn. If 600 feet of fencing is available to use, then what is the

area of the largest garden that could be built?
(A) Define a function that relates the area enclosed by the fence (in 〖ft.〗^2) in terms of its length (in ft.)
(B) What is the practical domain of this function?
(C) What is the largest area that the fence could enclose?
Mathematics
1 answer:
vovangra [49]3 years ago
4 0

Answer: (A) A=300l-l^{2}

               (B) Length varies between 1 and 150

               (C) Largest area is 22500ft²

Step-by-step explanation: Suppose length is l and width is w.

The rectangular garden has perimeter of 600ft, which is mathematically represented as

2l+2w=600

Area of a rectangle is calculated as

A=lw

Now, we have a system of equations:

2l+2w=600

A=lw

Isolate w, so we have l:

2w=600-2l

w = 300 - l

Substitute in the area equation:

A = l(300 - l)

A = 300l - l²

(A) <u>Function of area in terms of length is given by </u><u>A = 300l - l²</u>

(B) The practical domain for this function is values between 1 and 150.

(C) For the largest area, we need to determine the largest garden possible. For that, we take first derivative of the function:

A' = 300 - 2l

Find the values of l when A'=0:

300 - 2l = 0

2l = 300

l = 150

Replace l in the equation:

w = 300 - 150

w = 150

Now, calculate the largest area:

A = 150*150

A = 22500

<u>The largest area the fence can enclose is </u><u>22500ft².</u>

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b= - 15/6

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A real estate agent has 19 properties that she shows. She feels that there is a 30% chance of selling any one property during a
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Answer:

P(X \geq 5)=1-P(X

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P(X=0)=(19C0)(0.3)^0 (1-0.3)^{19-0}=0.00114

P(X=1)=(19C1)(0.3)^1 (1-0.3)^{19-1}=0.0092

P(X=2)=(19C2)(0.3)^2 (1-0.3)^{19-2}=0.0358

P(X=3)=(19C3)(0.3)^3 (1-0.3)^{19-3}=0.0869

P(X=4)=(19C4)(0.3)^4 (1-0.3)^{19-4}=0.1491

And replacing we got:

P(X \geq 5) = 1-[0.00114+0.009282+0.0358+0.0869+0.149]= 0.7178

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=19, p=0.3)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

And we want to find this probability:

P(X \geq 5)

And we can use the complement rule:

P(X \geq 5)=1-P(X

We can find the individual probabilities:

P(X=0)=(19C0)(0.3)^0 (1-0.3)^{19-0}=0.00114

P(X=1)=(19C1)(0.3)^1 (1-0.3)^{19-1}=0.0092

P(X=2)=(19C2)(0.3)^2 (1-0.3)^{19-2}=0.0358

P(X=3)=(19C3)(0.3)^3 (1-0.3)^{19-3}=0.0869

P(X=4)=(19C4)(0.3)^4 (1-0.3)^{19-4}=0.1491

And replacing we got:

P(X \geq 5) = 1-[0.00114+0.009282+0.0358+0.0869+0.149]= 0.7178

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3 years ago
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